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Dmitrij [34]
3 years ago
13

A certain shade of blue has a frequency of 7.28×1014 Hz.

Chemistry
1 answer:
Lyrx [107]3 years ago
7 0

Explanation:

the enerfy of of one photon of this light is 4.85x10^-19 J

<h2>E= 6.63 x 10^-34 J/S x 7.32 x10^14 S ^-1</h2>

E= 4.85x10^-19 J

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A sample of excited atoms lie 3.314×10−19 J above the ground state. Determine the emission wavelength of these atoms.
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The emission wavelength of the atoms excited above the ground state is 599.8 nm.

<h3>Emission wavelength of the atoms</h3>

The emission wavelength of the atom is determined by using the following formulas as shown below;

E = hf

E = hc/λ

where;

  • E is the energy of the atom
  • h is Planck's constant
  • c is speed of light
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λ = hc/E

λ = (6.626 x 10⁻³⁴ x 3 x 10⁸) / (3.314 x 10⁻¹⁹)

λ = 5.998 x 10⁻⁷ m

λ = 599.8 x 10⁻⁹ m

λ = 599.8 nm

Thus, the emission wavelength of the atoms is 599.8 nm.

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